NEET Oscillations Previous Year Questions with Solutions

Consider a cylinder of mass m and cross-sectional area A floating in equilibrium in a liquid of density rho. When depressed downward by distance y from equilibrium, additional buoyant force provides a restoring force:

• • Because restoring force is directly proportional to displacement y and directed toward equilibrium, the motion is Simple Harmonic Motion (SHM).
• • Effective restoring force constant: k = A * rho * g.

Apply the standard harmonic oscillator time period formula:

• • Mass m and area A are constant since the same cork is transferred between liquids.
• • Squaring gives: T^2 \propto \frac{1}{\rho} \implies \rho \propto \frac{1}{T^2}.

Substitute T_1 = T and T_2 = 2T into the inverse square relationship:

• • Confirms rho_2 / rho_1 = 1/4.
• • Matches Option (4) exactly.

For an undamped spring-mass oscillator of mass m and spring constant k, total mechanical energy E is conserved:

• • x is the extension/compression from equilibrium in meters.
• • v is the instantaneous velocity in m/s.

Divide both sides of the energy equation by total energy E:

• • This represents an ellipse on the x - v plane with horizontal semi-axis a = \sqrt{2E/k} and vertical semi-axis b = \sqrt{2E/m}.

An ellipse becomes a circle if and only if both semi-axes are equal in magnitude (a = b):

• • Equivalently, angular frequency omega = sqrt(k/m) = 1 rad/s.
• • Matches Option (2) exactly.

For a mass M attached to an ideal spring of stiffness k, the natural angular frequency of oscillation is:

• • As sand leaks slowly out of the box, M(t) decreases monotonically over time towards the empty box mass M_box.
• • Therefore, ω(t) increases monotonically, approaching a constant upper asymptote.

When sand leaks out vertically, the escaping grains have horizontal velocity equal to the instantaneous velocity v of the box at the moment of exit. The escaping sand exerts no horizontal force on the remaining box, but carries away its share of kinetic energy:

• • The total mechanical energy E(t) of the oscillator continuously decreases.
• • Since total energy is related to amplitude by E = (1/2) k A², we have A(t) = √(2 E(t) / k).
• • Consequently, as E(t) decreases, the average amplitude A(t) decreases with time.

The correct schematic shows ω(t) starting from initial value and curving upwards towards an asymptote, while A(t) starts from initial amplitude and curves downwards:

• • Matches Option (2) unambiguously.

For a particle executing simple harmonic motion with amplitude A and angular frequency ω, the maximum speed occurs at the equilibrium position:

• • Alternative derivation via energy conservation: (1/2) m v_{max}² = (1/2) k A² ⟹ v_{max} = A √(k/m).

Since the masses are identical and their maximum speeds are equal:

• • The common mass factor √m cancels out from both sides:
• • A_P \sqrt{k_1} = A_Q \sqrt{k_2}.

Rearrange the expression for the required ratio:

• • Matches Option (4) perfectly.

The general displacement equation of a particle executing simple harmonic motion (SHM) along the x-axis is given by:

Comparing the given equation x = 5 sin(πt + π/3) m with the standard form: - Amplitude A = 5 m - Angular frequency ω = π rad/s - Initial phase angle φ = π/3 rad

The time period T for one complete oscillation is related to the angular frequency ω by:

For small angular displacements, the time period T of a simple pendulum of length L and bob mass m oscillating under gravity g is given by:

Notice that the mass m of the bob does not appear in the formula. Restoring torque and rotational inertia both scale linearly with m, so mass cancels out identically. Therefore, tripling the mass does not change the time period at all.

The length is made half its original length: L’ = L / 2. The new time period T’ is:

We are given that T’ = (x / 2) T. Equating the two expressions:

The given sinusoidal waveform begins at x = 0 at t = 0, reaches its positive maximum x = +1 m at t = 2 s, returns to zero at t = 4 s, reaches its negative minimum x = -1 m at t = 6 s, and completes one full oscillation at t = 8 s:

The angular frequency of the simple harmonic oscillator is related to the time period T by:

Since the motion starts at the origin at t = 0 moving in the positive x direction, the displacement function is represented by:

Differentiating displacement twice with respect to time yields the acceleration function a(t) = -ω² x(t). At time t = 2 s, the displacement is x(2) = sin(π/2) = +1 m:

By Hooke's law, the restoring force is proportional to elongation: $F = kx$. Given $F = 10\text{ N}$ for extension $x = 5\text{ cm} = 5 \times 10^{-2}\text{ m}$:

For a block of mass $m = 2\text{ kg}$ oscillating on an ideal spring of stiffness $k$:

Substitute $\pi \approx 3.1416$:

For a particle executing simple harmonic motion with frequency $n$ (angular frequency $\omega = 2\pi n$) and amplitude $A$, displacement from the mean position is:

The potential energy $U$ of the harmonic oscillator with force constant $k$ is given by:

Using the identity $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$:

The time-dependent harmonic term $\cos(2\omega t)$ has angular frequency $\omega' = 2\omega$. Since $\omega = 2\pi n$:

Consider a particle executing simple harmonic motion about the origin. Its displacement as a function of time is given by: $$x(t) = A \sin(\omega t)$$ where $$A$$ is the amplitude and $$\omega$$ is the angular frequency. The initial phase is $$\phi_x = 0$$.

Velocity is the first time derivative of displacement: $$v(t) = \frac{dx}{dt} = A \omega \cos(\omega t) = A \omega \sin\left(\omega t + \frac{\pi}{2}\right)$$ Acceleration is the second time derivative of displacement: $$a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t)$$

Using the trigonometric identity $$-\sin(\theta) = \sin(\theta + \pi)$$: $$a(t) = A \omega^2 \sin(\omega t + \pi)$$ Here, the phase of acceleration is $$\phi_a = \omega t + \pi$$.

The phase difference between acceleration and displacement is: $$\Delta \phi = \phi_a - \phi_x = (\omega t + \pi) - (\omega t) = \pi\text{ rad} = 180^\circ$$ Physically, when displacement is at its maximum positive value ($$+A$$), acceleration is directed toward the center at its maximum negative value ($$-\omega^2 A$$).